Let $f(x)=\dfrac{\sqrt{x-1}-2}{x-5}$ when $x\neq 5$. $f$ is continuous for all $x>1$. Find $f(5)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{4}$ (Choice B) B $\dfrac{1}{10}$ (Choice C) C $\dfrac{1}{2}$ (Choice D) D $1$
Answer: $\dfrac{\sqrt{x-1}-2}{x-5}$ is continuous for all $x>1$ other than $x=5$, which means $f$ is continuous for all $x>1$ other than $x=5$. In order for $f$ to also be continuous at $x=5$, the following equality must hold: $\lim_{x\to 5}f(x)=f(5)$ We will obtain the above equality by letting $f(5)=\lim_{x\to 5}f(x)$. So let's find $\lim_{x\to 5}f(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to 5}f(x) \\\\ &=\lim_{x\to 5}\dfrac{\sqrt{x-1}-2}{x-5} \gray{\text{This is the rule for }x\neq 5} \\\\ &=\lim_{x\to 5}\dfrac{\sqrt{x-1}-2}{x-5}\cdot\dfrac{\sqrt{x-1}+2}{\sqrt{x-1}+2} \gray{\text{Rationalize}} \\\\ &=\lim_{x\to 5}\dfrac{x-1-2^2}{(x-5)(\sqrt{x-1}+2)} \gray{\text{Simplify}} \\\\ &=\lim_{x\to 5}\dfrac{\cancel{x-5}}{\cancel{(x-5)}(\sqrt{x-1}+2)} \gray{\text{Cancel common factors}} \\\\ &=\lim_{x\to 5}\dfrac{1}{(\sqrt{x-1}+2)} \\\\ &\text{(This is allowed because }x\neq 5) \\\\ &=\dfrac{1}{\sqrt{5-1}+2} \gray{\text{Direct substitution}} \\\\ &=\dfrac{1}{4} \end{aligned}$ We obtained that if we set $f(5)=\dfrac{1}{4}$, then $\lim_{x\to 5}f(x)=f(5)$, which makes $f$ continuous at $x=5$. Since we already saw that $f$ is continuous for any other $x>1$, we can determine that it's continuous for all $x>1$. In conclusion, $k=\dfrac{1}{4}$.